A nucleus of mass 218 amu is in free state decays to emit an `alpha`-particle. Kinetic energy of `alpha`-particle emitted is 6.7Mev. The recoil energy in (MeV) emitted by the daughter nucleus is

Options

A. `1.0`
B. `0.5`
C. 0.25
D. 0.125

Answer

To solve the problem, we will use the conservation of momentum and the relationship between kinetic energy and momentum.

### Step-by-Step Solution:

1. **Identify the Masses:**
- The mass of the original nucleus (before decay) is given as \( 218 \, \text{amu} \).
- The mass of the emitted alpha particle is \( 4 \, \text{amu} \).
- Therefore, the mass of the daughter nucleus after the emission of the alpha particle is:
\[ \text{Mass of daughter nucleus} = 218 \, \text{amu} - 4 \, \text{amu} = 214 \, \text{amu} \]

2. **Use Conservation of Momentum:**
- Since the nucleus is initially at rest, the total momentum before decay is zero. Thus, the momentum of the alpha particle must be equal in magnitude and opposite in direction to the momentum of the daughter nucleus.
- Let \( p_1 \) be the momentum of the alpha particle and \( p_2 \) be the momentum of the daughter nucleus:
\[ p_1 + p_2 = 0 \quad \Rightarrow \quad p_1 = -p_2 \]

3. **Relate Kinetic Energy and Momentum:**
- The kinetic energy \( K \) of a particle can be expressed in terms of its momentum \( p \) and mass \( m \):
\[ K = \frac{p^2}{2m} \]
- For the alpha particle:
\[ K_{\alpha} = \frac{p_1^2}{2m_{\alpha}} = 6.7 \, \text{MeV} \]
- For the daughter nucleus:
\[ K_d = \frac{p_2^2}{2m_d} \]

4. **Express \( p_2 \) in terms of \( p_1 \):**
- Since \( p_1 = -p_2 \), we have \( p_2 = -p_1 \) and thus:
\[ p_2^2 = p_1^2 \]

5. **Set Up the Energy Equation:**
- From the kinetic energy equations:
\[ \frac{p_1^2}{2m_{\alpha}} = 6.7 \, \text{MeV} \]
\[ \frac{p_1^2}{2m_d} = K_d \]
- Dividing these two equations gives:
\[ \frac{K_d}{6.7} = \frac{m_{\alpha}}{m_d} \]

6. **Substitute the Masses:**
- The mass of the alpha particle \( m_{\alpha} = 4 \, \text{amu} \) and the mass of the daughter nucleus \( m_d = 214 \, \text{amu} \):
\[ \frac{K_d}{6.7} = \frac{4}{214} \]

7. **Solve for \( K_d \):**
- Rearranging gives:
\[ K_d = 6.7 \times \frac{4}{214} \]
- Calculating:
\[ K_d = 6.7 \times \frac{4}{214} \approx 0.125 \, \text{MeV} \]

### Final Answer:
The recoil energy emitted by the daughter nucleus is approximately \( 0.125 \, \text{MeV} \).

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